How do you factor completely $25 a^{4} - 49 b^{2}$ $25a^4-49b^2$ ?

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How do you factor completely $25 a^{4} - 49 b^{2}$ $25a^4-49b^2$ ?

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Answer 1 · 2017-04-19T05:59:40

$25 a^{4} - 49 b^{2} = (5 a^{2} - 7 b) (5 a^{2} + 7 b)$ $25a^4-49b^2 = (5a^2-7b)(5a^2+7b)$

Explanation:

Note that both $25 a^{4} = {(5 a^{2})}^{2}$ $25a^4 = (5a^2)^2$ and $49 b^{2} = {(7 b)}^{2}$ $49b^2=(7b)^2$ are perfect squares.

The difference of squares identity can be written:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2=(A-B)(A+B)$

Using this with $A = 5 a^{2}$ $A=5a^2$ and $B = 7 b$ $B=7b$ we find:

$25 a^{4} - 49 b^{2} = {(5 a^{2})}^{2} - {(7 b)}^{2}$ $25a^4-49b^2 = (5a^2)^2-(7b)^2$

$25 a^{4} - 49 b^{2} = (5 a^{2} - 7 b) (5 a^{2} + 7 b)$ $color(white)(25a^4-49b^2) = (5a^2-7b)(5a^2+7b)$

There are no simpler factors.

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How do you factor completely $25 a^{4} - 49 b^{2}$ $25a^4-49b^2$ ?

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